Step-by-step Solution:

1. Start with the definition of conditional probability. We want to find the value of $P\left( {{E^c} \cap {F^c}|G} \right)$. The formula for conditional probability is $P(A|B) = \frac{P(A \cap B)}{P(B)}$. Applying this to our problem, we get: $P\left( {{E^c} \cap {F^c}|G} \right) = \frac{P\left( {{E^c} \cap {F^c} \cap G} \right)}{P(G)}$ We are given that $P(G) > 0$, so the denominator is non-zero.

2. Simplify the numerator using set theory and probability rules. First, we use De Morgan's Law on the term ${E^c} \cap {F^c}$, which states that ${E^c} \cap {F^c} = (E \cup F)^c$. So, the numerator becomes $P\left( {(E \cup F)^c \cap G} \right)$. Using the property $P(A^c \cap B) = P(B) - P(A \cap B)$, we can write: $P\left( {(E \cup F)^c \cap G} \right) = P(G) - P\left( {(E \cup F) \cap G} \right)$

3. Expand the term $P\left( {(E \cup F) \cap G} \right)$. Using the distributive property of set intersection over union, we have $(E \cup F) \cap G = (E \cap G) \cup (F \cap G)$. So, $P\left( {(E \cup F) \cap G} \right) = P\left( {(E \cap G) \cup (F \cap G)} \right)$. Now, we apply the principle of inclusion-exclusion for two events, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. $P\left( {(E \cap G) \cup (F \cap G)} \right) = P(E \cap G) + P(F \cap G) - P\left( {(E \cap G) \cap (F \cap G)} \right)$ The last term simplifies to $P(E \cap F \cap G)$. So, $P\left( {(E \cup F) \cap G} \right) = P(E \cap G) + P(F \cap G) - P(E \cap F \cap G)$

4. Use the given information about the events. We are given that $E, F, G$ are pairwise independent. This implies:

   • $P(E \cap G) = P(E)P(G)$
   • $P(F \cap G) = P(F)P(G)$ We are also given that $P(E \cap F \cap G) = 0$.

5. Substitute the given information into the expression. Substituting these into the equation from Step 3: $P\left( {(E \cup F) \cap G} \right) = P(E)P(G) + P(F)P(G) - 0$ $P\left( {(E \cup F) \cap G} \right) = P(G) \left( {P(E) + P(F)} \right)$

6. Calculate the numerator $P\left( {{E^c} \cap {F^c} \cap G} \right)$. From Step 2, we have $P\left( {(E \cup F)^c \cap G} \right) = P(G) - P\left( {(E \cup F) \cap G} \right)$. Substituting the result from Step 5: $P\left( {(E \cup F)^c \cap G} \right) = P(G) - P(G) \left( {P(E) + P(F)} \right)$ $P\left( {(E \cup F)^c \cap G} \right) = P(G) \left( {1 - P(E) - P(F)} \right)$

7. Calculate the final conditional probability. Now we substitute this back into the formula from Step 1: $P\left( {{E^c} \cap {F^c}|G} \right) = \frac{P(G) \left( {1 - P(E) - P(F)} \right)}{P(G)}$ Since $P(G) > 0$, we can cancel $P(G)$ from the numerator and denominator: $P\left( {{E^c} \cap {F^c}|G} \right) = 1 - P(E) - P(F)$

8. Match the result with the given options. We know that the probability of a complement event is $P(E^c) = 1 - P(E)$. Substituting this into our result: $P\left( {{E^c} \cap {F^c}|G} \right) = P(E^c) - P(F)$ This matches option C.

Alternative Approach:

Consider the probability space conditioned on G. Let $P_G(A) = P(A|G)$. We want to find $P_G(E^c \cap F^c)$.

1. Since E and G are independent, $P_G(E) = P(E|G) = \frac{P(E \cap G)}{P(G)} = \frac{P(E)P(G)}{P(G)} = P(E)$.
2. Similarly, since F and G are independent, $P_G(F) = P(F|G) = P(F)$.
3. Given $P(E \cap F \cap G) = 0$, we have $P_G(E \cap F) = P(E \cap F | G) = \frac{P(E \cap F \cap G)}{P(G)} = \frac{0}{P(G)} = 0$.
4. This means that in the conditioned space, events E and F are mutually exclusive.
5. We need to find $P_G(E^c \cap F^c) = P_G((E \cup F)^c) = 1 - P_G(E \cup F)$.
6. $P_G(E \cup F) = P_G(E) + P_G(F) - P_G(E \cap F) = P(E) + P(F) - 0 = P(E) + P(F)$.
7. Therefore, $P_G(E^c \cap F^c) = 1 - (P(E) + P(F)) = 1 - P(E) - P(F)$.
8. Rewriting this, we get $(1 - P(E)) - P(F) = P(E^c) - P(F)$. This confirms the previous result.